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ksaimouli Group Title

A 661-N force is being applied to a 121-kg block. The coefficient of static friction between the block and the surface is 0.410. What is the minimal amount of additional mass that must be added on top of the block to prevent the block from moving?

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  1. BTaylor Group Title
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    The frictional force is the coefficient of friction times the normal force. Here, you want the frictional force to equal 661, so there is no movement. So:\[F_{fr} = \mu_s \times F_N \rightarrow 661 = 0.410 \times mg\] From this, you can solve for mg, and you get mg = 1612 dividing by g (9.8), you get m = 164.5. So, since you already have 121 kg, you only need 43.5 more.

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